Question

The number of integral value(s) of $a$ for which the equation $2^a{x}^2-4^{a}x-2^a-1=0$ has exactly one root between $1$ and $2$ is

Answer 1

$2^a{x}^2-4^{a}x-2^a-1=0$
$\Rightarrow x^2-2^{a}x-1-\frac{1}{2^a}=0$
Let $f\left(x\right)=x^2-2^{a}x-1-\frac{1}{2^a}$
For exactly one root between $1$ and $2$
$f\left(1\right)f\left(2\right)<0$
$\Rightarrow (1-2^a-1-\frac{1}{2^a})(4-2\cdot 2^a-1-\frac{1}{2^a})<0$
Assuming $2^a=t$
$\Rightarrow (-1{)}^2(t+\frac{1}{t})(2t+\frac{1}{t}-3)<0\Rightarrow \frac{(t^2+1)(2t^2-3t+1)}{t^2}<0\Rightarrow \frac{(t^2+1)(2t-1)(t-1)}{t^2}<0\Rightarrow \frac{1}{2}<t<1\Rightarrow \frac{1}{2}<2^a<1$
Taking ${\log }_2$ on both sides,
$-1<a<0$

Checking boundary condition,
When $a=0$, we get
$x^2-x-2=0\Rightarrow x=-1,2$

When $a=-1$, we get
$\frac{x^2}{2}-\frac{x}{4}-\frac{1}{2}-1=0\Rightarrow 2x^2-x-6=0\Rightarrow (2x+3)(x-2)=0\Rightarrow x=-\frac{3}{2},2$

Hence, $a\in (-1,0)$, so no integral value of $a$ lies in this interval.